Theorem. A field extension $L/K$ is algebraic if and only if any subring of the extension field $L$ containing the base field $K$ is a field.

Proof. Assume first that $L/K$ is algebraic. Let $R$ be a subring of $L$ containing $K$ . For any non-zero element $r$ of $R$ , naturally $K[r] \subseteq R$ , and since $r$ is an algebraic element over $K$ , the ring $K[r]$ coincides with the field $K(r)$ . Therefore we have $r^{-1} \in K[r] \subseteq R$ , and $R$ must be a field.

Assume then that each subring of $L$ which contains $K$ is a field. Let $a$ be any non-zero element of $L$ . Accordingly, the subring $K[a]$ of $L$ contains $K$ and is a field. So we have $a^{-1} \in K[a]$ . This means that there is a polynomial $f(x)$ in the polynomial ring $K[x]$ such that $a^{-1} = f(a)$ . Because $af(a)-1 = 0$ , the element $a$ is a zero of the polynomial $xf(x)-1$ of $K[x]$ , i.e. is algebraic over $K$ . Thus every element of $L$ is algebraic over $K$ .

Bibliography

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DAVID M. BURTON: A first course in rings and ideals. Addison-Wesley Publishing Company. Reading, Menlo Park, London, Don Mills (1970).